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What a fun question! The intuitive answer is that it would be very hard to detect any half-life difference because the electrons are smeared out over something like the Bohr radius, while the nucleus is much smaller, so the contribution of the electron charge to the problem is negligible. However if the correct radius to consider is the distance over which the alpha particles "tunnel" to escape from the nucleus, the effect might be quite substantial for highly-charged ions.

Let's consider the tunneling model for alpha decay. We'll assume a nucleus with $Z$ protons and radius $a \approx A^{1/3}\times1.3\,\rm fm$. Inside the nucleus, thanks to the strong interaction, the alpha particle sees some constant potential that we don't really care about. Outside the nucleus, the alpha particle is electrically repelled. So the total potential, in spherical coordinates, is $$ V(r) = \left\{ \begin{array}{cc} V_0 & r < a \\ \displaystyle \alpha \hbar c \frac{2(Z-2)}{r} & r > a \end{array} \right. $$ This potential totally ignores all the atomic electrons.

The electrons will contribute a potential $$ V_\text{e}(r) = -\alpha\hbar c \frac{2q_\text{enc}(r)}r $$ where $q_\text{enc}$ is the charge (in units of the charge quantum $e$) enclosed within the radius $r$. (The shell theorem for electromagnetism says that we can safely ignore charges which are uniformly distributed at large radius.) The spherical harmonics with $\ell\neq0$ vanish at the origin, so we need only consider the $s$-wave electrons. The radial hydrogenic wavefunctions with $\ell=0$ have a uniform probability density near the origin of $$ \left|\psi(r=0)\right|^2 = \frac1{\pi a_e^3} = \frac1\pi \left( \frac{Z}{na_0} \right)^3 $$ where $a_0 \approx 53\,000\,\rm fm$ is the Bohr radius. This gives us, for an electron in the $n$th $s$ shell, a charge enclosed within radius $r$ of $$ q_\text{enc}^n(r) = \frac 43 \left( \frac Zn \frac{r}{a_0} \right)^3. $$ Let's try it for a $1s$ electron in uranium (largest $Z$), and compare the nuclear charge $Z$ to the charge of electrons which are visiting, using the definition of nuclear radius given above: \begin{align} q_\text{enc}^1 (a_{\text{U-238}}) &= \frac43 \left( \frac{92}1 \frac{1.3\,\rm fm}{a_0} \right)^3 A \tag1 \\&\approx 4\times10^{-6} \end{align} However, the nuclear volume isn't quite the right thing to consider here. The alpha particle escapes the nucleus by tunneling through the barrier. The distance where the alpha's wavefunction changes from decaying exponential to sinusoidal is where the final kinetic energy is equal to the electrostatic energy, \begin{align} E_\alpha &= V(r) \\ E_\alpha &= \alpha \hbar c \frac{2(Z-2)}{r} \\ \text{5 MeV from uranium: } r &= \frac{200\rm\,MeV\,fm}{137} \frac{2\cdot90}{5\rm\,MeV} \approx 52\rm\,fm. \end{align} The relation (1) above suggests that this larger volume contains about $0.23e$ from each of the two $1s$ electrons, $0.03e$ from each of the two $2s$ electrons, and the rest don't matter so much. This suggests that the width of Coulomb barrier might be different at the 0.5% level for completely ionized uranium compared to partially ionized uranium (since the $1s$ electrons will be the last to go). The decay rate difference between different ionization states sounds modest but experimentally accessible to me, but your question is the first I've heard of it.

What a fun question! The intuitive answer is that it would be very hard to detect any half-life difference because the electrons are smeared out over something like the Bohr radius, while the nucleus is much smaller, so the contribution of the electron charge to the problem is negligible. However if the correct radius to consider is the distance over which the alpha particles "tunnel" to escape from the nucleus, the effect might be quite substantial for highly-charged ions.

Let's consider the tunneling model for alpha decay. We'll assume a nucleus with $Z$ protons and radius $a \approx A^{1/3}\times1.3\,\rm fm$. Inside the nucleus, thanks to the strong interaction, the alpha particle sees some constant potential that we don't really care about. Outside the nucleus, the alpha particle is electrically repelled. So the total potential, in spherical coordinates, is $$ V(r) = \left\{ \begin{array}{cc} V_0 & r < a \\ \displaystyle \alpha \hbar c \frac{2(Z-2)}{r} & r > a \end{array} \right. $$ This potential totally ignores all the atomic electrons.

The electrons will contribute a potential $$ V_\text{e}(r) = -\alpha\hbar c \frac{2q_\text{enc}(r)}r $$ where $q_\text{enc}$ is the charge (in units of the charge quantum $e$) enclosed within the radius $r$. (The shell theorem for electromagnetism says that we can safely ignore charges which are uniformly distributed at large radius.) The spherical harmonics with $\ell\neq0$ vanish at the origin, so we need only consider the $s$-wave electrons. The radial hydrogenic wavefunctions with $\ell=0$ have a uniform probability density near the origin of $$ \left|\psi(r=0)\right|^2 = \frac1{\pi a_e^3} = \frac1\pi \left( \frac{Z}{na_0} \right)^3 $$ where $a_0 \approx 53\,000\,\rm fm$ is the Bohr radius. This gives us, for an electron in the $n$th $s$ shell, a charge enclosed within radius $r$ of $$ q_\text{enc}^n(r) = \frac 43 \left( \frac Zn \frac{r}{a_0} \right)^3. $$ Let's try it for a $1s$ electron in uranium (largest $Z$), and compare the nuclear charge $Z$ to the charge of electrons which are visiting, using the definition of nuclear radius given above: \begin{align} q_\text{enc}^1 (a_{\text{U-238}}) &= \frac43 \left( \frac{92}1 \frac{1.3\,\rm fm}{a_0} \right)^3 A \tag1 \\&\approx 4\times10^{-6} \end{align} However, the nuclear volume isn't quite the right thing to consider here. The alpha particle escapes the nucleus by tunneling through the barrier. The distance where the alpha's wavefunction changes from decaying exponential to sinusoidal is where the final kinetic energy is equal to the electrostatic energy, \begin{align} E_\alpha &= V(r) \\ E_\alpha &= \alpha \hbar c \frac{2(Z-2)}{r} \\ \text{5 MeV from uranium: } r &= \frac{200\rm\,MeV\,fm}{137} \frac{2\cdot90}{5\rm\,MeV} \approx 52\rm\,fm. \end{align} The relation (1) above suggests that this larger volume contains about $0.23e$ from each of the two $1s$ electrons, $0.03e$ from each of the two $2s$ electrons, and the rest don't matter so much. This suggests that the width of Coulomb barrier might be different at the 0.5% level for completely ionized uranium compared to partially ionized uranium (since the $1s$ electrons will be the last to go). The decay rate difference between different ionization states sounds modest but experimentally accessible to me, but your question is the first I've heard of it.

A recent preprint by F. Belloni, with a nice bibliography, studies this phenomenon in much more detail. I neglected to include the change in the $Q$-value of the decay between the bare nucleus and the neutral ion, which also affects the and makes the whole question rather complicated. Belloni computes a lifetime difference of 0.1% between bare Po-210 and hydrogen-like Po-210 and larger changes in α-lifetimes at extremely high electron densities. In general Belloni predicts shorter α-decay lifetimes in the presence of electrons.

What a fun question! The answer is that it would be very hard to detect any half-life difference because the electrons are smeared out over something like the Bohr radius, while the nucleus is much smaller, so the contribution of the electron charge to the problem is negligible.

Let's consider the tunneling model for alpha decay. We'll assume a nucleus with $Z$ protons and radius $a \approx A^{1/3}\times1.3\,\rm fm$. Inside the nucleus, thanks to the strong interaction, the alpha particle sees some constant potential that we don't really care about. Outside the nucleus, the alpha particle is electrically repelled. So the total potential, in spherical coordinates, is $$ V(r) = \left\{ \begin{array}{cc} V_0 & r < a \\ \displaystyle \alpha \hbar c \frac{2(Z-2)}{r} & r > a \end{array} \right. $$ This potential totally ignores all the atomic electrons.

The electrons will contribute a potential $$ V_\text{e}(r) = -\alpha\hbar c \frac{2q_\text{enc}(r)}r $$ where $q_\text{enc}$ is the charge (in units of the charge quantum $e$) enclosed within the radius $r$. (The shell theorem for electromagnetism says that we can safely ignore charges which are uniformly distributed at large radius.) The spherical harmonics with $\ell\neq0$ vanish at the origin, so we need only consider the $s$-wave electrons. The radial hydrogenic wavefunctions with $\ell=0$ have a uniform probability density near the origin of $$ \left|\psi(r=0)\right|^2 = \frac1{\pi a_e^3} = \frac1\pi \left( \frac{Z}{na_0} \right)^3 $$ where $a_0 \approx 53\,000\,\rm fm$ is the Bohr radius. This gives us, for an electron in the $n$th $s$ shell, a charge enclosed within radius $r$ of $$ q_\text{enc}^n(r) = \frac 43 \left( \frac Zn \frac{r}{a_0} \right)^3. $$ Let's try it for a $1s$ electron in uranium (largest $Z$), and compare the nuclear charge $Z$ to the charge of electrons which are visiting, using the definition of nuclear radius given above: $$ q_\text{enc}^1 (a_\text{U}) = \frac43 \left( \frac{92}1 \frac{1.3\,\rm fm}{a_0} \right)^3 A $$ It doesn't really matter what we plug in for $n,Z,A$: the ratio of the length scales $(1.3\,\rm fm /53\,000\,fm)^3$ means there will roughly a billionth of a negative charge quantum inside the nucleus for the alpha particle to interact with.

What a fun question! The intuitive answer is that it would be very hard to detect any half-life difference because the electrons are smeared out over something like the Bohr radius, while the nucleus is much smaller, so the contribution of the electron charge to the problem is negligible. However if the correct radius to consider is the distance over which the alpha particles "tunnel" to escape from the nucleus, the effect might be quite substantial for highly-charged ions.

Let's consider the tunneling model for alpha decay. We'll assume a nucleus with $Z$ protons and radius $a \approx A^{1/3}\times1.3\,\rm fm$. Inside the nucleus, thanks to the strong interaction, the alpha particle sees some constant potential that we don't really care about. Outside the nucleus, the alpha particle is electrically repelled. So the total potential, in spherical coordinates, is $$ V(r) = \left\{ \begin{array}{cc} V_0 & r < a \\ \displaystyle \alpha \hbar c \frac{2(Z-2)}{r} & r > a \end{array} \right. $$ This potential totally ignores all the atomic electrons.

The electrons will contribute a potential $$ V_\text{e}(r) = -\alpha\hbar c \frac{2q_\text{enc}(r)}r $$ where $q_\text{enc}$ is the charge (in units of the charge quantum $e$) enclosed within the radius $r$. (The shell theorem for electromagnetism says that we can safely ignore charges which are uniformly distributed at large radius.) The spherical harmonics with $\ell\neq0$ vanish at the origin, so we need only consider the $s$-wave electrons. The radial hydrogenic wavefunctions with $\ell=0$ have a uniform probability density near the origin of $$ \left|\psi(r=0)\right|^2 = \frac1{\pi a_e^3} = \frac1\pi \left( \frac{Z}{na_0} \right)^3 $$ where $a_0 \approx 53\,000\,\rm fm$ is the Bohr radius. This gives us, for an electron in the $n$th $s$ shell, a charge enclosed within radius $r$ of $$ q_\text{enc}^n(r) = \frac 43 \left( \frac Zn \frac{r}{a_0} \right)^3. $$ Let's try it for a $1s$ electron in uranium (largest $Z$), and compare the nuclear charge $Z$ to the charge of electrons which are visiting, using the definition of nuclear radius given above: \begin{align} q_\text{enc}^1 (a_{\text{U-238}}) &= \frac43 \left( \frac{92}1 \frac{1.3\,\rm fm}{a_0} \right)^3 A \tag1 \\&\approx 4\times10^{-6} \end{align} However, the nuclear volume isn't quite the right thing to consider here. The alpha particle escapes the nucleus by tunneling through the barrier. The distance where the alpha's wavefunction changes from decaying exponential to sinusoidal is where the final kinetic energy is equal to the electrostatic energy, \begin{align} E_\alpha &= V(r) \\ E_\alpha &= \alpha \hbar c \frac{2(Z-2)}{r} \\ \text{5 MeV from uranium: } r &= \frac{200\rm\,MeV\,fm}{137} \frac{2\cdot90}{5\rm\,MeV} \approx 52\rm\,fm. \end{align} The relation (1) above suggests that this larger volume contains about $0.23e$ from each of the two $1s$ electrons, $0.03e$ from each of the two $2s$ electrons, and the rest don't matter so much. This suggests that the width of Coulomb barrier might be different at the 0.5% level for completely ionized uranium compared to partially ionized uranium (since the $1s$ electrons will be the last to go). The decay rate difference between different ionization states sounds modest but experimentally accessible to me, but your question is the first I've heard of it.

What a fun question! The answer is that it would be very hard to detect any half-life difference because the electrons are smeared out over something like the Bohr radius, while the nucleus is much smaller, so the contribution of the electron charge to the problem is negligible. (Edited to add: the extended "nuclear" volume probed by the tunneling alpha particle is still very much smaller than the Bohr radius.)

Let's consider the tunneling model for alpha decay. We'll assume a nucleus with $Z$ protons and radius $a \approx A^{1/3}\times1.3\,\rm fm$. Inside the nucleus, thanks to the strong interaction, the alpha particle sees some constant potential that we don't really care about. Outside the nucleus, the alpha particle is electrically repelled. So the total potential, in spherical coordinates, is $$ V(r) = \left\{ \begin{array}{cc} V_0 & r < a \\ \displaystyle \alpha \hbar c \frac{2(Z-2)}{r} & r > a \end{array} \right. $$ This potential totally ignores all the atomic electrons.

The electrons will contribute a potential $$ V_\text{e}(r) = -\alpha\hbar c \frac{2q_\text{enc}(r)}r $$ where $q_\text{enc}$ is the charge (in units of the charge quantum $e$) enclosed within the radius $r$. (The shell theorem for electromagnetism says that we can safely ignore charges which are uniformly distributed at large radius.) The spherical harmonics with $\ell\neq0$ vanish at the origin, so we need only consider the $s$-wave electrons. The radial hydrogenic wavefunctions with $\ell=0$ have a uniform probability density near the origin of $$ \left|\psi(r=0)\right|^2 = \frac1{\pi a_e^3} = \frac1\pi \left( \frac{Z}{na_0} \right)^3 $$ where $a_0 \approx 53\,000\,\rm fm$ is the Bohr radius. This gives us, for an electron in the $n$th $s$ shell, a charge enclosed within radius $r$ of $$ q_\text{enc}^n(r) = \frac 43 \left( \frac Zn \frac{r}{a_0} \right)^3. $$ Let's try it for a $1s$ electron in uranium (largest $Z$), and compare the nuclear charge $Z$ to the charge of electrons which are visiting, using the definition of nuclear radius given above: $$ q_\text{enc}^1 (a_\text{U}) = \frac43 \left( \frac{92}1 \frac{1.3\,\rm fm}{a_0} \right)^3 A \tag 1$$ It doesn't really matter what we plug in for $n,Z,A$: the ratio of the length scales $(1.3\,\rm fm /53\,000\,fm)^3$ means there will roughly a billionth of a negative charge quantum inside the nucleus for the alpha particle to interact with.

Edited: Actually, the nuclear volume is the wrong volume to consider here. Instead we should use the length through which the alpha particle must tunnel to escape the nucleus. If the alpha's final kinetic energy is $E_\alpha$, its wavefunction switches from exponential to sinusoidal when \begin{align} E_\alpha &= V(r) \\ E_\alpha &= \alpha \hbar c \frac{2(Z-2)}{r} \end{align} For a 5 MeV alpha from uranium this gives a tunneling radius of $$ r = \frac{200\,\rm MeV\,fm}{137} \frac{2 \cdot 90}{5\,\rm MeV} = 50\,\rm fm , $$ and within a volume of this radius, (1) gives a few parts per thousand of an electron charge quantum mostly from the $1s$ and $2s$ electrons. This suggests you might actually see a (very) slightly reduced half-life in heavy alpha-emitters missing their $1s$ electrons. The fractional effect of the electron charge is going to be bigger for low-energy decays, where the tunneling radius is longer; however the low-energy decays are already the slower ones, which is where the alpha-tunneling model came from in the first place.

Of course an atom with its $1s$ electrons missing is only electronically stable if it's stripped to a bare nucleus. This suggests a very challenging experiment, comparing decay rates (not necessarily energies, I think?) between neutral atoms and highly-charged ions, over a fairly long time, to a precision of a few parts per thousand. I doubt it's been done, but I'll update if I find anything.

What a fun question! The answer is that it would be very hard to detect any half-life difference because the electrons are smeared out over something like the Bohr radius, while the nucleus is much smaller, so the contribution of the electron charge to the problem is negligible.

Let's consider the tunneling model for alpha decay. We'll assume a nucleus with $Z$ protons and radius $a \approx A^{1/3}\times1.3\,\rm fm$. Inside the nucleus, thanks to the strong interaction, the alpha particle sees some constant potential that we don't really care about. Outside the nucleus, the alpha particle is electrically repelled. So the total potential, in spherical coordinates, is $$ V(r) = \left\{ \begin{array}{cc} V_0 & r < a \\ \displaystyle \alpha \hbar c \frac{2(Z-2)}{r} & r > a \end{array} \right. $$ This potential totally ignores all the atomic electrons.

The electrons will contribute a potential $$ V_\text{e}(r) = -\alpha\hbar c \frac{2q_\text{enc}(r)}r $$ where $q_\text{enc}$ is the charge (in units of the charge quantum $e$) enclosed within the radius $r$. (The shell theorem for electromagnetism says that we can safely ignore charges which are uniformly distributed at large radius.) The spherical harmonics with $\ell\neq0$ vanish at the origin, so we need only consider the $s$-wave electrons. The radial hydrogenic wavefunctions with $\ell=0$ have a uniform probability density near the origin of $$ \left|\psi(r=0)\right|^2 = \frac1{\pi a_e^3} = \frac1\pi \left( \frac{Z}{na_0} \right)^3 $$ where $a_0 \approx 53\,000\,\rm fm$ is the Bohr radius. This gives us, for an electron in the $n$th $s$ shell, a charge enclosed within radius $r$ of $$ q_\text{enc}^n(r) = \frac 43 \left( \frac Zn \frac{r}{a_0} \right)^3. $$ Let's try it for a $1s$ electron in uranium (largest $Z$), and compare the nuclear charge $Z$ to the charge of electrons which are visiting, using the definition of nuclear radius given above: $$ q_\text{enc}^1 (a_\text{U}) = \frac43 \left( \frac{92}1 \frac{1.3\,\rm fm}{a_0} \right)^3 A $$ It doesn't really matter what we plug in for $n,Z,A$: the ratio of the length scales $(1.3\,\rm fm /53\,000\,fm)^3$ means there will roughly a billionth of a negative charge quantum inside the nucleus for the alpha particle to interact with.